# What is the definition of Pure Mathematics?

Oh, easy, you say it is just the study of numbers.

That may be true for some areas of math. However, what about geometry, trigonometry, and calculus? And what is the definition of numbers? Now you go to the dictionary and say The relationship between measurements and quantities using numbers and symbols. This is, however, not fully true because this definition also uses applied mathematics. We want pure mathematics.

Also, most of these definitions miss one area of math. Chaos Theory. What is Chaos Theory? Chaos Theory is a recently discovered area of math where nothing can be predicted but nothing is random. We are only at the beginning of learning it. For example, can a butterfly that flaps its wings in Brazil trigger a tornado in Texas?

Some definitions hit almost all the areas of math, but some are too broad and logic, for example, often fits into the definition.

We can, however, define some areas of math but not the whole thing. For example, the definition of geometry is Geometry is concerned with the properties and relations of points, lines, surfaces, solids, and higher dimensional analogs. Or the definition of probability is the extent to which an event is likely to occur.

# Arithmetic

## = Definition

The branch of mathematics dealing with the properties and manipulation of constants(a number that can not changed. This is the oppisite of a varible which is most of the tie repersented by a letter. This letter is most of the time $x$,$y$ or $z$ however it can be really any non-used symbol.).

### Operations

Arithmetic starts with one thing which without it no arithmetic can survive: Counting Positive Integers (whole numbers). 1,2,3,4,5...

Addition is combining these integers. The symbol for combining numbers is +. $a+b=b+a$

Multiplication is repeating addition. The symbol for this is $\cdot$. $a \cdot b=b \cdot a$. Note: When using letters, you can just say $ab=ba$.

Exponentiation is repeated Multiplication. The symbol for a to the exponent of b is $a^b$. $a^b \neq b^a$. The $\neq$ symbol means Not Equal.

#### Inverse

Subtraction is the inverse (remember that inverse means oppisite) of addition. To make this a well-defined function (the symbol for a funtion of a certian number($x$) is $f(x)=$blah blah blah where a function is a relationship or expression involving one or more variables. For example, if $f(x)=x+3$ then $f(3)=6$), negative numbers and zero were defined. A negitive number is a number below zero.

The division is the inverse of multiplication. To make this a well-defined function everywhere, fractions were defined. 3 divided by 2 is $\frac{3}{2}$.

$n^{\text{th}}$ roots and logarithms are the inverses of exponentiation. To make these well-defined functions everywhere, irrational numbers were defined. An irrational number is a number that can not be expressed as $\frac{a}{b}. ==== Negative numbers ====$ (Error compiling LaTeX. ! Missing $inserted.)a$and$b$are positive. (Above zero)

1.$(Error compiling LaTeX. ! Missing$ inserted.)(-a)(-b)=ab$Note: 2.$ (Error compiling LaTeX. ! Missing $inserted.)(a)(-b)=-ab$Proof for 1: This is, in fact, the reason why the negative numbers were introduced: so that each positive number would have an additive inverse. ... The fact that the product of two negatives is positive is therefore related to the fact that the inverse of the inverse of a positive number is that positive number back again.

Proof for 2: Since ab is repeated addition then$(Error compiling LaTeX. ! Missing$ inserted.)(a)(-b)$is repeated subtraction. Therefore it is negative. ==== Fractions ==== A fraction a number that can be expressed as two numbers divided. For example, five divided by four is$ (Error compiling LaTeX. ! Missing $inserted.)\frac{5}{4}$. ===== Simplifing fractions ===== Find common factors in each half of the fration and then divide top and bottam of the fraction by that factor.$\frac{10}{5}=\frac{1}{2}$$(Error compiling LaTeX. ! Missing inserted.)\frac{15}{3}=\frac{5}{1}$$ (Error compiling LaTeX. ! Missing$ inserted.)\frac{28}{2}=\frac{14}{1}$===== Adding fractions ===== Here is how you add fractions. Is they have the same bottom half then$\frac{a}{b}+\frac{c}{b}=\frac{a+c}{b}$. However, if the two or three or n fractions do not have the same bottom half you make them.$\frac{a}{b}+\frac{c}{d}=\frac{ad+bc}{bd}$. Subtracting fractions is the same exept everything has a minus symbol. Remember to fully simplify. ===== Multilpying fractions =====$ (Error compiling LaTeX. ! Missing $inserted.)\frac{a}{b}\frac{c}{d}=\frac{ac}{bd}$.$\frac{\frac{a}{b}}{\frac{c}{d}}=\frac{ac}{bd}$.

Remember to fully simplify.

==== Irrational numbers ==== Most square roots are irrational. An irrational number is a number that can not be expresed a a fraction. ===== Proof that any nonperfect square positive integer is irrational ===== Let us assume that$(Error compiling LaTeX. ! Missing$ inserted.)\sqrt{n}$is rational where$n$is a nonperfect square positive integer. Then it can be written as$\frac{p}{q}=\sqrt{n} \Rightarrow \frac{p^2}{q^2}=n \Rightarrow (q^2)n=p^2$. But no perfect square times a nonperfect square positive integer is a perfect square. Therefore$\sqrt{n}$is irrational. ===== pi and e ===== Some irrational numbers are limits. That means they are the sum of smaller and smaller fraction going infinitely long. Pi or$ (Error compiling LaTeX. ! Missing $inserted.)\pi$(go to the geometry part of this article) is the limit$\frac{1}{1}-\frac{1}{3}+\frac{1}{5}-\frac{1}{7}...=\frac{\pi}{4}$. e is the limit$\frac{1}{0!}+\frac{1}{1!}+\frac{1}{2!}...$More at the counting part of the article.

=== Exponent rules === Listed below are some important properties of exponents:

1. $(Error compiling LaTeX. ! Missing$ inserted.) b^x\cdot b^y = b^{x+y}$#$ (Error compiling LaTeX. ! You can't use macro parameter character #' in math mode.) b^{-x}=\frac 1{b^x} $#$ (Error compiling LaTeX. ! You can't use macro parameter character #' in math mode.) \frac{b^x}{b^y}=b^{x-y} $#$ (Error compiling LaTeX. ! You can't use macro parameter character #' in math mode.) (b^x)^y = b^{xy} $#$ (Error compiling LaTeX. ! You can't use macro parameter character #' in math mode.) (ab)^x = a^x b^x $#$ (Error compiling LaTeX. ! You can't use macro parameter character #' in math mode.) b^0 = 1 $(if$b \neq 0$.$0^0$is undefined.) Here are explanations of the properties listed above: 1. On both sides, we are multiplying '''b''' together '''x+y''' times. Thus, they are equivalent. 2. This is described in the previous section. 3. This results from using the previous two properties. 4. We are multiplying$ (Error compiling LaTeX. ! Missing $inserted.)b^x$by itself '''y''' times, which is the same as multiplying '''b''' by itself '''xy''' times.
5. After multiplying '''ab''' by itself '''x''' times, we can collect '''a''' and '''b''' terms, thus establishing the property.
6. Hoping that property #1 will be true when$(Error compiling LaTeX. ! You can't use macro parameter character #' in math mode.)y=0$, we see that$b^x\cdot b^0$should (hopefully) be equal to$b^x$. Thus, we ''define''$b^0$to be equal to$1$in order to make this be true.

= Algebra = == Definition == ''The part of mathematics in which letters and other general symbols are used to represent numbers and quantities in formulae and equations.'' == One-variable linear equations == === Definition === 'A One-variable linear equation is an equation that comes in the form$(Error compiling LaTeX. ! Missing$ inserted.)ax+b=c$.$a$,$b$, and$c$are constants and$x$is the varible in which to solve for === Problems === ==== Problem 1 ==== Solve for$x$:$x+3=7$==== Solution 1 ==== Subtract 3 from both sides you get$x=4$==== Problem 2 ==== Solve for$x$:$3x+2=98$==== Solution 2 ==== Subtract two:$3x=96$. Divide by three:$x=32$==== Problem 3 ==== Solve for$x$:$\sqrt{x+3}=5$==== Solution 3 ==== When dealing with square roots you just square roots you square them because that makes no more square roots. So$\sqrt{x+3}=5$squared is$x+3=25$which tells us that$x$is$22$. ==== Problem 4 ==== Solve for$x$:$\frac{4}{x}=8$==== Solution 4 ==== When dealing with fractions you multiply by the denaminator:$4=8x$which tells us that$x=\frac{1}{2}$. == Quadratics == === Defination === ''A quadratic equation is an equation of the second degree, meaning it contains at least one term that is squared. The standard form is$ (Error compiling LaTeX. ! Missing $inserted.)ax^2 + bx + c = 0$with$a$,$b$, and$c$being constants, or numerical coefficients, and$x$is an unknown variable.''

=== The answer is always... ===$(Error compiling LaTeX. ! Missing$ inserted.)ax^2+bx+c=0$$(Error compiling LaTeX. ! Missing  inserted.)x^2+\frac{b}{a}x+\frac{c}{a}=0$$ (Error compiling LaTeX. ! Missing $inserted.)x^2+\frac{b}{a}=0-\frac{c}{a}$$(Error compiling LaTeX. ! Missing inserted.)x^2+\frac{b}{a}+(\frac{b}{2a})^2=0-\frac{c}{a}+(\frac{b}{2a})^2$Since the left side of the equation right above is a perfect square, you can factor the left side by using the coefficient of the first term (x) and the base of the last term(b/2a). Add these two and raise everything to the second.$(x+\frac{b}{2a})^2=0-\frac{c}{a}+\frac{b^2}{2a^2}=\frac{b^2-4ac}{4a^2}$$ (Error compiling LaTeX. ! Missing$ inserted.)x+\frac{b}{2a}=\frac{\sqrt{\pm b^2-4ac}}{\pm 2a}$$(Error compiling LaTeX. ! Missing  inserted.)x=\frac{\sqrt{\pm b^2-4ac}}{\pm 2a}+\frac{b}{2a}$When simplified the ''Quadratic Formula'' is${x=\frac{-b \pm \sqrt{b^2-4ac}}{2a}}$===== Problem ===== Find all$x$if$x^2+x+1=0$===== Solution ===== Using our formula we see that$x=\frac{-1 \pm \sqrt{1^2-1(1)(1)}}{2(1)}=-\frac{1}{2}$== i ==$i=\sqrt{-1}$.$xi=\sqrt{-x^2}. Numbers like this are called ''imaginary numbers''. ''Impossible'', you say. But no. Solve (Error compiling LaTeX. ! Missing  inserted.)x^2+1=0$. You get$i$.$oi$is$o. So zero is both real and imaginary. (real means ''not imaginary'') === Powers of i === (Error compiling LaTeX. ! Missing  inserted.)i^0=1$$ (Error compiling LaTeX. ! Missing $inserted.)i^1=i$$(Error compiling LaTeX. ! Missing inserted.)i^2=-1$$ (Error compiling LaTeX. ! Missing$ inserted.)i^3=-i$$(Error compiling LaTeX. ! Missing  inserted.)i^4=1$$ (Error compiling LaTeX. ! Missing $inserted.)i^5=i$$(Error compiling LaTeX. ! Missing inserted.)i^6=-1$$ (Error compiling LaTeX. ! Missing$ inserted.)i^7=-i$The pattern repeats. === Complex numbers === A complex number is$ (Error compiling LaTeX. ! Missing $inserted.)ai+b$, where a and b are real. All numbers are complex becuase a and/or/never b can be zero.$ai+b+ci+d=$Complex$(ai+b)(ci+d)=$Complex$\frac{ai+b}{ci+d}=$Complex$(ai+b)-(ci+d)=$Complex

== Systems of equations == A '''system of equations''' is a set of [[equation]]s which share the same [[variable]]s. An example of a system of equations is

{| class="wikitable" style="margin: 1em auto 1em auto" |$(Error compiling LaTeX. ! Missing$ inserted.)2a - 3b$||$= 4$|- |$3a - 2b$||$= 3$|} === Solving Linear Systems === A system of [[linear]] equations is where all of the variables are to the power 1. There are three elementary ways to solve a system of linear equations. ==== Gaussian Elimination ==== [[Gauss]]ian elimination involves eliminating variables from the system by adding constant multiples of two or more of the equations together. Let's look at an example: ===== Problem ===== Find the ordered pair$ (Error compiling LaTeX. ! Extra }, or forgotten $.)(x,y)$for which

{| class="wikitable" style="margin: 1em auto 1em auto" |$(Error compiling LaTeX. ! Missing$ inserted.)x - 12y$||$= 2$|- |$3x + 6y$||$= 6$|} ===== Solution ===== We can eliminate$ (Error compiling LaTeX. ! Extra }, or forgotten $.)y$by adding twice the second equation to the first:

{| class="wikitable" style="margin: 1em auto 1em auto" | ||$(Error compiling LaTeX. ! Missing$ inserted.)x - 12y= 2$|- |$+2($||$3x + 6y = 6)$|- | ||$\overline{7x + 0=14}$|} Thus$ (Error compiling LaTeX. ! Extra }, or forgotten $.)x=2$. We can then plug in for$x$in either of the equations: $ (2)-12y = 2 \Rightarrow y = 0$.</center>

Thus, the solution to the system is$(Error compiling LaTeX. ! Missing$ inserted.)(2,0)$. ==== Substitution ==== The second method, substitution, requires solving for a variable and then plugging that variable into another equation therefore reducing the number of variables. We'll show how to solve the same problem from the elimination section using substitution. ===== Problem ===== Find the ordered pair$ (Error compiling LaTeX. ! Missing $inserted.)(x,y)$for which

{| class="wikitable" style="margin: 1em auto 1em auto" |$(Error compiling LaTeX. ! Missing$ inserted.)x - 12y$||$= 2$|- |$3x + 6y$||$= 6$|} ===== Solution ===== The first equation can be solved for$ (Error compiling LaTeX. ! Extra }, or forgotten $.)x$: $ x = 12y + 2.$</center>

Plugging this into the second equation yields <center>$(Error compiling LaTeX. ! Missing$ inserted.)3(12y + 2) + 6y = 6 \Leftrightarrow 42 y = 0. $</center> Thus$ (Error compiling LaTeX. ! Missing $inserted.)y=0$. Plugging this into either of the equations and solving for$x$yields$x=2$.

== Graphing Equations == == Graphing Lines == Given two distinct points on a line, one can construct the whole line. So one way to graph a line given its equation is to just find two points on it and to draw a straight line through them.

=== Problem === Graph the line$(Error compiling LaTeX. ! Missing$ inserted.)2x + 3y = 24$. === Solution === To graph a line, it is necesasry to find two points$ (Error compiling LaTeX. ! Missing $inserted.)(x,y)$that satisfy$2x + 3y = 24$. Letting$x=0$gives$3y = 24\Leftrightarrow y = 8$. So$(0,8)$is one point on the graph.

Find another point by letting$(Error compiling LaTeX. ! Missing$ inserted.)y=0$. Plugging this in and solving gives$x=12$. So$(12,0)$is our other point. Now plot these in the coordinate plane and draw a line through them: <center>[[Image:Twopoints2.PNG]]</center> The arrowheads on the ends of the line segment indicate that the line goes on [[infinite]]ly in both directions. Note: This is just a short intro to graphing. More at https://artofproblemsolving.com/wiki/index.php/Graph_of_a_function. == Algebra == There are many more types of algebra: inequalities, polynomials, graphing equations, arithmetic, and geometric sequence. Algebra is a broad and diverse area of math in which this is just a short introduction. = Number Theory = == Definition == ''The branch of mathematics that deals with the properties and relationships of numbers, especially the positive integers''. == Vocab == Factor: A factor (f) of n is an integer in which$ (Error compiling LaTeX. ! Missing $inserted.)fx=n$where$x$is an integer.

Multiple: The inverse of factors.

== Primes == A '''prime number''' (or simply '''prime''') is a positive integer$(Error compiling LaTeX. ! Missing$ inserted.)p>1$whose only positive divisors are 1 and itself. Note that$1$is usually defined as being neither prime nor composite because it is its only factor among the natural number numbers. There are an infinite number of prime numbers. A standard proof attributed to Euclid notes that if there are a finite set of prime numbers$ (Error compiling LaTeX. ! Missing $inserted.)p_1, p_2, \ldots, p_n$, then the number$N = p_1p_2\cdots p_n + 1$is not divisible by any of them, but$N$must [[#Importance of Primes|have]] a prime factor, which leads to a direct contradiction.

=== Techniques to Check for Prime Numbers ===

==== Divisibility ==== A prime number is only divisible by one or itself, so a number$(Error compiling LaTeX. ! You can't use macro parameter character #' in math mode.)n$is prime if and only if$n$is not divisible by any integer greater than$1$and less than$n$. One only needs to check integers up to$\sqrt{n}$because dividing larger numbers would result in a quotient smaller than$\sqrt{n}$.

==== Modular Arithmetic ==== Modular arithmetic can help determine if a number is not prime.

• If a number not equal to$(Error compiling LaTeX. ! Missing$ inserted.)2,3$is congruent to$0,2,3,4,6 \pmod{6}$, then the number is not prime. * If a number not equal to$2,5$ends with an even digit or$5$, then the number is not prime. === Importance of Primes === According to the Fundamental Theorem of Arithmetic, there is exactly one unique way to factor a positive integer into a product of primes. This unique prime factorization plays an important role in solving many kinds of number theory problems. ==== Mersenne Primes ==== A Mersenne prime is a prime of the form$ (Error compiling LaTeX. ! Missing $inserted.)2^n-1$. For such a number to be prime, ''n'' must itself be prime. Compared to other numbers of comparable sizes, Mersenne numbers are easy to check for primality because of the https://en.wikipedia.org/wiki/Lucas%E2%80%93Lehmer_primality_test Lucas-Lehmer test, severely reducing the amount of computation needed.

==== Twin Primes ==== Two primes that differ by exactly 2 are known as twin primes. The following are the first few pairs of twin primes:<br> 3, 5<br> 5, 7<br> 11, 13<br> 17, 19<br> 29, 31<br> 41, 43<br>

It is not known whether or not there are infinitely many pairs of twin primes. This is known as the Twin Prime Conjecture, which is a specific instance of the Hardy-Littlewood conjecture.

==== Gaussian Primes ====

A Gaussian prime is a prime that extends the idea of the traditional prime to the Gaussian integer. One can define this term for any ring, especially number rings.

=== Advanced Definition === When the need arises to include negative divisors, a '''prime''' is defined as an integer p whose only divisors are 1, -1, p, and -p. More generally, if ''R'' is an integral domain, then a nonzero element ''p'' of ''R'' is a '''prime''' if whenever we write$(Error compiling LaTeX. ! Missing$ inserted.)p=ab$with$a,b\in R$, then exactly one of ''a'' and ''b'' is a unit. == LCM and GCD == === GCD === ==== Definition ==== The GCD or Greatest Common Divisor of multiple numbers is the largest number that is a factor of all those multiple numbers. The answer can be and usually is 1. ==== Prime Factorization ==== The way to solve most GCD's is with using ''Prime Factorization''. Remember that the way to find the prime factorization of a number is to find which primes products hit that certain number. The prime factorization of 24 is$ (Error compiling LaTeX. ! Missing $inserted.)2 \cdot 2 \cdot 2 \cdot 3$. The prime factorization of 18 is$2 \cdot 3 \cdot 3$. In GCD you take the prime factorization of each number and then find which primes match up.

=== LCM === ==== Definition ==== The LCM or Least Common Multiple of two or three or n numbers is the smallest integer in which both numbers are factors of it. For example, the LCM of 10 and 5 is 10. ==== Problems ==== ===== Problem 1 ===== ====== Problem ====== What is the LCM of 8 and 5? ====== Solution ====== The way to solve LCM's with no common factors is to multiply them. The answer is just 40. ===== Problem 2 ===== ====== Problem ====== What is the LCM of 15 and 10. ====== Solution ====== The Way to solve this is see that the answer would be the first multiple of 5 that is divisble by 15 and 10. That is just 30. ===== Problem 3 ===== ====== Problem ====== Find the LCM of 2, 3 and 5. ====== Solution ====== It is the same with three numbers. Multiply them to get 30. Same as last time.

== Bases == In mathematics, a base or radix is the number of different digits or a combination of digits and letters that a system of counting uses to represent numbers. For example, the most common base used today is the decimal system. Because "dec" means 10, it uses the 10 digits from 0 to 9. Most people think that we most often use base 10 because we have 10 fingers. ==== Some Numbers in other bases ==== A base can be any whole number bigger than 1. The base of a number may be written next to the number: for instance,$(Error compiling LaTeX. ! Missing$ inserted.){23_{8}}$means 23 in base 8 (which is equal to 19 in base 10). Here are some examples of how numbers are written in varying bases, compared to decimal: In Arabic numbers (decimal, or base 10), there are 10 digits: 0,1,2,3,4,5,6,7,8,9. You need one digit each to count up to 9, but two digits for ten, and three digits for a hundred, which is ten times ten. In Binary, base 2, you need two digits for two, as you only have two digits, 0 and 1. Base 5 has five digits, and the number five becomes 10. For base 16, you will need sixteen digits, and there are only ten numerals. So we use the letters A,B,C,D,E,F. These represent the decimal numbers 10, 11, 12, 13, 14 and 15. Look at the table below and find the pattern for these bases. == Number Theory == This is the end of the number theory part of this article. Number Theory, however, has much more to than what I have just shown you. = Probibility = == Definition == ''The extent to which an event is likely to occur.'' == How to == The way to mesure the extent to which an event is likely to occur is to count the number of equally likely occur is to count the wanted options over the number of equally likelly possible options. For example, the probibility of rolling a 6 on a 6 sided dice is$ (Error compiling LaTeX. ! Missing $inserted.)\frac{1}{6}$because there are six equally likely possibilitys and only one of them is a success.

= Pascal's Triangle = == Pascal's Identity == === Identity === Pascal's Identity states that$(Error compiling LaTeX. ! Missing$ inserted.){n \choose k}={n-1\choose k-1}+{n-1\choose k}$for any positive integers$k$and$n$. Here,$\binom{n}{k}$is the binomial coefficient$\binom{n}{k} = nCk = C_k^n$. Remember that$ (Error compiling LaTeX. ! Missing $inserted.)\binom{n}{r}=\frac{n!}{k!(n-k)!}$.$\binom{n}{r}$means the number of ways to pick r thing from n things where order does not matter.$n!=1 \cdot 2 \cdot 3 \cdot...\cdot n$.

=== Proving it === If$(Error compiling LaTeX. ! Missing$ inserted.)k > n$then$\binom{n}{k} = 0 = \binom{n - 1}{k - 1} + \binom{n - 1}{k}$and so the result is pretty clear. So assume$k \leq n$. Then <cmath>\begin{eqnarray*}\binom{n-1}{k-1}+\binom{n-1}{k}&=&\frac{(n-1)!}{(k-1)!(n-k)!}+\frac{(n-1)!}{k!(n-k-1)!}\\ &=&(n-1)!\left(\frac{k}{k!(n-k)!}+\frac{n-k}{k!(n-k)!}\right)\\ &=&(n-1)!\cdot \frac{n}{k!(n-k)!}\\ &=&\frac{n!}{k!(n-k)!}\\ &=&\binom{n}{k}. \qquad\qquad\square\end{eqnarray*}</cmath> There we go. We proved it! === Why is it needed? === It's mostly just a cool thing to know. However, if you want to know how to use it in real life go to https://artofproblemsolving.com/videos/counting/chapter12/141. Or really any of the counting and probability videos. == Introduction to Pascal's Triangle == === How to build it === Pascal's Triangle is a triangular array of numbers in which you start with two infinite diagonals of ones and each of the rest of the numbers is the sum of the two numbers above it. It looks something like this:  1 1 1 1 2 1 1 3 3 1 1 4 6 4 1  And on and on... === Combinations === ==== Combinations ==== Pascal's Triangle is really combinations. It looks something like this if it is depicted as combinations:$ (Error compiling LaTeX. ! Missing $inserted.)\binom{0}{0}$$(Error compiling LaTeX. ! Missing inserted.)\binom{1}{0}$$ (Error compiling LaTeX. ! Missing$ inserted.)\binom{1}{1}$$(Error compiling LaTeX. ! Missing  inserted.)\binom{2}{0}$$ (Error compiling LaTeX. ! Missing $inserted.)\binom{2}{1}$$(Error compiling LaTeX. ! Missing inserted.)\binom{2}{2}And on and on... ==== Proof ==== If you look at the way we build the triangle, each number is the sum of the two numbers above it. Assuming that these combinations are true then each combination in the sum of the two combinations above it. In an equation, it would look something like this: (Error compiling LaTeX. ! Missing inserted.){n \choose k}={n-1\choose k-1}+{n-1\choose k}$. Its Pascals Identity! Therefore each row looks something like this:$\binom{n}{0} \binom{n}{1} \binom{n}{2} ... \binom{n}{n}$== Patterns and Properties == In addition to combinations, ''Pascal's Triangle'' has many more patterns and properties. See below. Be ready to be amazed. === Binomial Theorem === Let's multiply out some binomials. Try it yourself and it will not be fun:$(x+y)^0=1$$ (Error compiling LaTeX. ! Missing$ inserted.)(x+y)^1=1x+1y$$(Error compiling LaTeX. ! Missing  inserted.)(x+y)^2=1x^2+2xy+1y^2$$ (Error compiling LaTeX. ! Missing $inserted.)(x+y)^2=1x^3+3x^2y+3y^2x+1^3$If you take away the x's and y's you get:

         1
1 1
1 2 1
1 3 3 1
`

It's ''Pascal's Triangle''!

===== Proof ===== Here are a number of different ways to prove the Binomial Theorem, for example by a straightforward application of mathematical induction. The Binomial Theorem also has a nice combinatorial proof:

We can write$(Error compiling LaTeX. ! Missing$ inserted.)(a+b)^n=\underbrace{ (a+b)\cdot(a+b)\cdot(a+b)\cdot\cdots\cdot(a+b) }_{n}$. Repeatedly using the distributive property, we see that for a term$a^m b^{n-m}$, we must choose$m$of the$n$terms to contribute an$a$to the term, and then each of the other$n-m$terms of the product must contribute a$b$. Thus, the coefficient of$a^m b^{n-m}$is the number of ways to choose$m$objects from a set of size$n$, or$\binom{n}{m}$. Extending this to all possible values of$m$from$0$to$n$, we see that$(a+b)^n = \sum_{m=0}^{n}{\binom{n}{m}}\cdot a^m\cdot b^{n-m}$, as claimed. Similarly, the coefficients of$ (Error compiling LaTeX. ! Missing $inserted.)(x+y)^n$will be the entries of the$n^\text{th}$row of Pascal's Triangle. This is explained further in the Counting and Probability textbook [AoPS] ===== In real life ===== It is really only used for multipling out binomials. More usage at https://artofproblemsolving.com/videos/counting/chapter14/126. ==== Powers of 2 ==== ===== Theorem ===== ====== Theorem ====== It states that$\binom{n}{0}+\binom{n}{1}+...+\binom{n}{n}=2^n$====== Why do we need it? ====== It is useful in many word problems (That means, yes, you can use it in real life) and it is just a cool thing to know. More at https://artofproblemsolving.com/videos/mathcounts/mc2010/419. ===== Proofs ===== ====== Subset proof ====== Say you have a word with n letters. How many subsets does it have in terms of n? Here is how you answer it: You ask the first letter Are you in or are you out? Same to the second letter. Same to the third. Same to the n. Each of the letters has two choices: In and Out. The would be$(2)(2)(2)(2)$...n times.$2^n$.

====== Alternate proof ====== If you look at the way we built the triangle you see that each number is row n-1 is added on twice in row n. This means that each row doubles. That means you get powers of two.

==== Triangle Numbers ==== ===== Theorem ===== If you look at the numbers in the third diagonal you see that they are triangle numbers. ===== Proof ===== Now we can make an equation:$(Error compiling LaTeX. ! Missing$ inserted.)\binom{n}{2}=1+2+3+...+(n-1) \Rightarrow \binom{n}{2}=\frac{n(n+1)}{2} \Rightarrow \frac{n!}{2!(n-2)!}=\frac{n(n+1)}{2} \Rightarrow \frac{n(n+1)}{2}=\frac{n(n+1)}{2}$==== Hockey stick ==== For$n,r\in\mathbb{N}, n>r,\sum^n_{i=r}{i\choose r}={n+1\choose r+1}\$.

$[asy] int chew(int n,int r){ int res=1; for(int i=0;i

This identity is known as the hockey-stick identity because, on Pascal's triangle, when the addends represented in the summation and the sum itself is highlighted, a hockey-stick shape is revealed.

##### Proof

Inductive Proof

This identity can be proven by induction on $n$.

Base Case Let $n=r$.

$\sum^n_{i=r}{i\choose r}=\sum^r_{i=r}{i\choose r}={r\choose r}=1={r+1\choose r+1}$.

Inductive Step Suppose, for some $k\in\mathbb{N}, k>r$, $\sum^k_{i=r}{i\choose r}={k+1\choose r+1}$. Then $\sum^{k+1}_{i=r}{i\choose r}=\left(\sum^k_{i=r}{i\choose r}\right)+{k+1\choose r}={k+1\choose r+1}+{k+1\choose r}={k+2\choose r+1}$.

Algebraic Proof

It can also be proven algebraically with Pascal's Identity, ${n \choose k}={n-1\choose k-1}+{n-1\choose k}$. Note that

${r \choose r}+{r+1 \choose r}+{r+2 \choose r}+\cdots+{r+a \choose r}$ $={r+1 \choose r+1}+{r+1 \choose r}+{r+2 \choose r}+\cdots+{r+a \choose r}$ $={r+2 \choose r+1}+{r+2 \choose r}+\cdots+{r+a \choose r}=\cdots={r+a \choose r+1}+{r+a \choose r}={r+a+1 \choose r+1}$, which is equivalent to the desired result.

Combinatorial Proof 1

Imagine that we are distributing $n$ indistinguishable candies to $k$ distinguishable children. By a direct application of Balls and Urns, there are ${n+k-1\choose k-1}$ ways to do this. Alternatively, we can first give $0\le i\le n$ candies to the oldest child so that we are essentially giving $n-i$ candies to $k-1$ kids and again, with Balls and Urns, ${n+k-1\choose k-1}=\sum_{i=0}^n{n+k-2-i\choose k-2}$, which simplifies to the desired result.

Combinatorial Proof 2

We can form a committee of size $k+1$ from a group of $n+1$ people in ${{n+1}\choose{k+1}}$ ways. Now we hand out the numbers $1,2,3,\dots,n-k+1$ to $n-k+1$ of the $n+1$ people. We can divide this into $n-k+1$ disjoint cases. In general, in case $x$, $1\le x\le n-k+1$, person $x$ is on the committee and persons $1,2,3,\dots, x-1$ are not on the committee. This can be done in $\binom{n-x+1}{k}$ ways. Now we can sum the values of these $n-k+1$ disjoint cases, getting $${{n+1}\choose {k+1}} ={{n}\choose{k}}+{{n-1}\choose{k}}+{{n-2}\choose{k}}+\hdots+{{k+1}\choose{k}}+{{k}\choose{k}}.$$

# Geometry

## Studying the Geometry of Shapes

### Zero-D

This is an infinitly small dot called a point. (Duh). However, mathematicians imagine them to be large enough to be drawn.

### One-D

1D is pretty boring too. You just have rays, lines, and line segments.

#### Line

A infinitly long, infinitly narrow, perfectly striaght thing in with no starting or ending point. (I cant really use the word lines).

#### Ray

A infinitly long, infinitly narrow, perfectly striaght thing in with one starting point.

#### Line segment

A infinitly long, infinitly narrow, perfectly striaght thing in with two starting points.

### Plane

In mathematics, a plane is a flat, two-dimensional surface that extends infinitely far.

#### 3D Space

Space is the infinitly three-dimensional extent in which objects and events have relative position and direction.

### Angles

An angle is the figure formed by two rays, called the sides of the angle, sharing a common endpoint, called the vertex of the angle. Angles formed by two rays lie in a plane, but this plane does not have to be a Euclidean plane. Angles are also formed by the intersection of two planes in Euclidean and other spaces. These are called dihedral angles. Angles formed by the intersection of two curves in a plane are defined as the angle determined by the tangent rays at the point of intersection. Similar statements hold in space, for example, the spherical angle formed by two great circles on a sphere is the dihedral angle between the planes determined by the great circles. Most people mesure angles useing degrees. Degrees are out of 360. For example, 180° or 180 degrees is a straight line or a half turn.

An angle equal to 0° is called a zero angle.

An angle smaller than 90° but not zero is called acute.

An angle equal to 90° is called right.

An angle greater than 90° but less than 180° is called obtuse.

An angle equal to 180° is called striaght.

An angle greater than 180° but less than 360° is called obtuse.

An angle equal to 360° is called a full angle.

Angles $n$ and $m$ supplementary if $n+m=$180° Angles $n$ and $m$ Complementary if $n+m=$90°

Vertical angles: Each of the pairs of opposite angles made by two intersecting lines.

## Area

### Area

The area of a 2D object is the amount of square units in it.

### Shapes

#### Polygons

A plane shape (two-dimensional) with straight sides.

Examples: triangles, rectangles and pentagons.

(Note: a circle is not a polygon because it has a curved side).

#### Triangles and their classification

##### Angle Classifacation

Acute Triangle All angles are less than 90°

Right Triangle Has a right angle (90°)

Obtuse Triangle Has an angle more than 90°

##### Side Classifacation

Equilateral Triangle Three equal sides

Isosceles Triangle Two equal sides Two equal angles

Scalene Triangle No equal sides No equal angles

This part shows what Proves ussually look like

## Proof that any nonperfect square positive integer is irrational

Let us assume that $\sqrt{n}$ is rational where $n$ is a nonperfect square positive integer. Then it can be written as $\frac{p}{q}=\sqrt{n} \Rightarrow \frac{p^2}{q^2}=n \Rightarrow (q^2)n=p^2$. But no perfect square times a nonperfect square positive integer is a perfect square. Therefore $\sqrt{n}$ is irrational.

## Proof of the Pythagorean Theorem

$ABCD$ and $EFGH$ are squares.

$[asy] pair A, B,C,D; A = (-10,10); B = (10,10); C = (10,-10); D = (-10,-10); pair E,F,G,H; E = (7,10); F = (10, -7); G = (-7, -10); H = (-10, 7); draw(A--B--C--D--cycle); label("A", A, NNW); label("B", B, ENE); label("C", C, ESE); label("D", D, SSW); draw(E--F--G--H--cycle); label("E", E, N); label("F", F,SE); label("G", G, S); label("H", H, W); label("a", A--B,N); label("a", B--F,SE); label("a", C--G,S); label("a", H--D,W); label("b", E--B,N); label("b", F--C,SE); label("b", G--D,S); label("b", A--H,W); label("c", E--H,NW); label("c", E--F); label("c", F--G,SE); label("c", G--H,SW); [/asy]$

$(a+b)^2=c^2+4\left(\frac{1}{2}ab\right)\implies a^2+2ab+b^2=c^2+2ab\implies a^2 + b^2=c^2$.

## Proof that n choose 2 is 1+2+3...+(n-1)

$\binom{n}{2}=1+2+3...+n \Rightarrow \binom{n}{2}=\frac{n(n-1)}{2} \Rightarrow \frac{n(n-1)}{2}=\frac{n(n-1)}{2}$.Yay, I proved it!

## Proof that $1+2+3...+n+1+2+3...+(n-1)=n^2$

Proof that $1+2+3...+n+1+2+3...+(n-1)=n^2$

### Proof 1

$1+2+3...+n+1+2+3...+(n-1)=n^2 \Rightarrow \frac{n(n+1)}{2}+\frac{n-1(n)}{2}=n^2 \Rightarrow \frac{n(n+1)+n-1(n)}{2}=n^2 \Rightarrow \frac{n^2+n+n^2-n}{2}=n^2 \Rightarrow \frac{2n^2}{2}=n^2 \Rightarrow n^2=n^2$.

### Proof 2

The $1+2+\cdots+n$ part refers to an $n$ by $n$ square cut by its diagonal, and includes all the squares on the diagonal. The $1+2+\cdots+ n-1$ part refers to an $n$ by $n$ square cut by its diagonal, but doesn't include the squares on the diagonal. Putting these together gives us a $n$ by $n$ square.

### Proof 3

We proceed using induction. If $n = 1$, then we have $1+0=1^2$. Now assume that $n$ works. We prove that $n+1$ works. We add a $2n+1$ on both sides, such that the left side becomes $1+2+\cdots + (n+1)+1+2+\cdots + n = n^2 + 2n + 1 = (n+1)^2$ and we are done.

### Proof 4

$1 2 3 4 5 ... n$

$0 1 2 3 4 ... (n-1)$

________________

$1 3 5 7 9 ... 2n-1$ And that is $n^2$.